One atom of gold weighs 3.27 × 10^{-13} ng. How many gold atoms are present in 1.0 g of gold?

• A. 3.1 × 10^{15}
• B. 3.1 × 10^{21}
• C. 3.1 × 10^{12}
• D. 3.1 × 10^{18}

Answer: (A)

Verified step by step guidance

1. Convert the mass of one gold atom from nanograms (ng) to grams (g) to ensure consistent units. Since 1 ng = 10^{-9} g, multiply the given mass by 10^{-9}: \$3.27 \times 10^{-13} \text{ ng} = 3.27 \times 10^{-13} \times 10^{-9} \text{ g}$.
2. Set up the relationship to find the number of atoms in 1.0 g of gold by dividing the total mass of gold by the mass of one gold atom: \(\text{Number of atoms} = \frac{1.0 \text{ g}}{\text{mass of one atom in g}}\).
3. Substitute the converted mass of one atom into the equation: \(\text{Number of atoms} = \frac{1.0}{3.27 \times 10^{-22}}\) (using the result from step 1).
4. Perform the division by applying the rules of exponents to simplify the expression without calculating the exact number: \(\frac{1.0}{a \times 10^{b}} = \frac{1.0}{a} \times 10^{-b}\).
5. Express the final answer in scientific notation, matching it to one of the provided options by comparing the order of magnitude and coefficient.